2. The two diagrams to the right il-lustrate plots of steady-state ki-netic studies to characterize the in-teraction of heart muscle phos-phofructokinase-1 with a non-phy-siological, synthetic substrate fruc-tose-6-sulfate. Because the kcat issmaller than that for the natural50.8-NH =3.50.6-0.4-0.2-10 μΜ20 μΜ48 µMsubstrate, higher enzyme concen-trations could be used. The resultsshow the influence of increasing20.2-0.4-concentrations of ATP on the initial-0.6->velocity of the enzyme catalyzedreaction in the presence of no-0.8F41220283644526068768492.22.02.4ΑMP () , 10 μΜ AMP (+ ) 20 μΜAMP (-), and 48 µM AMP (-).[AΤP] (μΜ)log[ATP] (µM)(a) .Write the reaction in words catalyzed by the enzyme for the alternative substrate, describehow ATP interacts with the enzyme in the case of no AMP (•).(b) !with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme.Explain the physical significance of the displacement of the Hill plots to the right in panel B10 (µmoles/min/mg)A (A - "A) boj

Phosphofructokinase adds a phosphoryl group from ATP to Fructose 6 phosphate (F6P) to yield Fructose…

Answered: 2. The two diagrams to the right il-…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Which substrate is the best substrate for chymotrypsin? O N-acetylphenylalanine ethyl ester with Km of 8.2 x 102 M and keat of 2.1 x 102 sec 1. O N-acetylmethionine ethyl ester with Km of 12.3 x 102 M and keat of 2.1 x 1o sec1. O N-acetylleucine ethyl ester with Km of 8.2 x 10o2 M and kcat of 4.1 x 102 sec1. O N-acetylalanine ethyl ester with Km of 4.1 x 101 M and kcat of 4.1 x 101 sec.
TABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate…
In a molecular disease of hemoglobin, Hemoglobin Rainier, Tyr 145β is replaced by Cys, which forms a disulfide bond with another Cysresidue in the same subunit. This prevents the formation of ion pairs that normally stabilize the T state. How does hemoglobin Rainier differ from normal hemoglobin with respect to the Hill coefficient
Antithrombin III forms an irreversible complex with thrombin but not with prothrombin. What is the most likely reason for this difference in reactivity?
A binding curve for the binding of the amino acid tryptophan to the protein called TxtE is shown below. Estimate the association equilibrium constant (Ka; in units of M-1) for Trp binding to TxtE.
Could you please help answer this question?
. Studies on CAMP actions in cultured cells usually involve adding to the cell culture not CAMP, but dibutyryl CAMP (see structure). Why is this structural modification necessary? How could you test the premise that di-Bu-CAMP has the same biochemical effects as CAMP? N. N= Nat Dibutyryl CAMP O
PTP1B Substrate kcat Km. kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEP YLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 + 0.22 0.53 + 0.02 34.7 + 0.25 52.7 + 0.7 0.066 + 0.001 ) The units for kcat/KM in the above are given according to standard scientific notation. On this (d) ( basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
distribution of Benzene in a 4-compartment PBK model after an exposure of 6h. Convert Berkeley Madonna code to Matlab. Note: The parameters in the table below are for a mouse model. 1a) What are the two type of products containing Benzene consumers were recently concerned about. Which of these two different type of products is likely more harmful? (hint: check news/reports and social media channels) 1b) Plot Jresp along with the concentrations of Benzene in blood, fat, non-fat and liver in a 24h window (pdf). Why is Jresp slightly decreasing within 6h of exposure? Here is the Berkeley Madonna code: {Top model} METHOD RK4 STARTTIME = 0 STOPTIME = 24 DT = 0.02 {Reservoirs Blood, Fat, NonFat and Liver} d/dt (NF) = + Jnf INIT NF = 0 d/dt (F) = + Jf INIT F = 0 d/dt (B) = - Jnf - Jf - Jl + Jresp INIT B = 0 d/dt (L) = + Jl - Jmetab INIT L = 0 {Flows} Jnf = Qnf*(Cb-Cnf) Jf = Qf*(Cb-Cfv) Jl = Ql*(Cb - Clv) Jmetab = vmax*Cl/(Km +…
Calculate KI' of the inhibitor from the information given. All information may not be needed to calculate. K'm = (29Ki+1.45x10^-10)/Ki Vmax = 11.7 µMs-1 Kcat = 130 s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM Inhibitor Concentration = 5x10^-12
PTP1B Substrate kcat Km kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEPYLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 ± 0.22 0.53 + 0.02 34.7 + 0.25 52.7 ± 0.7 0.066 + 0.001 (d) (. ) The units for kcat/KM in the above are given according to standard scientific notation. On this basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
Calculate the concentration of p-aminophenol control to match the 0.075 % w/w limit in 200 mg/10 ml Your Answer: 0.0015% show work

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